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edit approved May 25, 2015 at 19:22
User12345
edit approved May 25, 2015 at 19:22
User12345
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In my assignment I have to prove that the following function is discontinuous:

$$f(x)= (2x-1) \space if \space x\notin \Bbb Q $$ $$f(x)=x^2 \space if \space x \in \Bbb Q$$$$f(x)=\begin{cases}2x-1&\text{if }x\notin\Bbb Q\\x^2&\text{if }x \in \Bbb Q\end{cases}$$

I have to prove that in terms of $\epsilon$ and $\delta$

I have some sort of beginning, I have to prove that some $\epsilon >0$ exists such that for every $\delta>0$ if $|x-x_0|<\delta \to |f(x)-f(x_0)| >\ \epsilon$

I don't really how to continue from here - I thought about picking $\epsilon=0.01$ and show that $|2x-4|$ is always bigger than that if I pick the right $\delta$ - but how do I show it? I have thought that since $\Bbb Q$ is a Dense set I can always choose the $\delta$ I want but... I don't know how.

Thanks,

Alan

In my assignment I have to prove that the following function is discontinuous:

$$f(x)= (2x-1) \space if \space x\notin \Bbb Q $$ $$f(x)=x^2 \space if \space x \in \Bbb Q$$

I have to prove that in terms of $\epsilon$ and $\delta$

I have some sort of beginning, I have to prove that some $\epsilon >0$ exists such that for every $\delta>0$ if $|x-x_0|<\delta \to |f(x)-f(x_0)| >\ \epsilon$

I don't really how to continue from here - I thought about picking $\epsilon=0.01$ and show that $|2x-4|$ is always bigger than that if I pick the right $\delta$ - but how do I show it? I have thought that since $\Bbb Q$ is a Dense set I can always choose the $\delta$ I want but... I don't know how.

Thanks,

Alan

In my assignment I have to prove that the following function is discontinuous:

$$f(x)=\begin{cases}2x-1&\text{if }x\notin\Bbb Q\\x^2&\text{if }x \in \Bbb Q\end{cases}$$

I have to prove that in terms of $\epsilon$ and $\delta$

I have some sort of beginning, I have to prove that some $\epsilon >0$ exists such that for every $\delta>0$ if $|x-x_0|<\delta \to |f(x)-f(x_0)| >\ \epsilon$

I don't really how to continue from here - I thought about picking $\epsilon=0.01$ and show that $|2x-4|$ is always bigger than that if I pick the right $\delta$ - but how do I show it? I have thought that since $\Bbb Q$ is a Dense set I can always choose the $\delta$ I want but... I don't know how.

Thanks,

Alan

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Alan
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In my assignment I have to prove that the following function is discontinuous:

$$f(x)= (2x-1) \space if \space x\notin \Bbb Q $$ $$f(x)=x^2 \space if \space x \in \Bbb Q$$

I have to prove that in terms of $\epsilon$ and $\delta$

I have some sort of beginning, I have to prove that some $\epsilon >0$ exists such that for every $\delta>0$ if $|x-x_0|<\delta \to |f(x)-f(x_0)| >\ \epsilon$

I don't really how to continue from here - I thought about picking $\epsilon=0.01$ and show that $|2x-4|$ is always bigger than that if I pick the right $\delta$ - but how do I show it? I have thought that since $\Bbb Q$ is a Dense set I can always choose the $\delta$ I want but... I don't know how.

Thanks,

Alan

In my assignment I have to prove that the following function is discontinuous:

$$f(x)= (2x-1) \space if \space x\notin \Bbb Q $$ $$f(x)=x^2 \space if \space x \in \Bbb Q$$

I have to prove that in terms of $\epsilon$ and $\delta$

I have some sort of beginning, I have to prove that some $\epsilon >0$ exists such that for every $\delta>0$ if $|x-x_0|<\delta \to |f(x)-f(x_0)| >\ \epsilon$

I don't really how to continue from here - I thought about picking $\epsilon=0.01$ and show that $|2x-4|$ is always bigger than that if I pick the right $\delta$ - but how do I show it?

Thanks,

Alan

In my assignment I have to prove that the following function is discontinuous:

$$f(x)= (2x-1) \space if \space x\notin \Bbb Q $$ $$f(x)=x^2 \space if \space x \in \Bbb Q$$

I have to prove that in terms of $\epsilon$ and $\delta$

I have some sort of beginning, I have to prove that some $\epsilon >0$ exists such that for every $\delta>0$ if $|x-x_0|<\delta \to |f(x)-f(x_0)| >\ \epsilon$

I don't really how to continue from here - I thought about picking $\epsilon=0.01$ and show that $|2x-4|$ is always bigger than that if I pick the right $\delta$ - but how do I show it? I have thought that since $\Bbb Q$ is a Dense set I can always choose the $\delta$ I want but... I don't know how.

Thanks,

Alan

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Alan
Alan
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Proving that a function is discontinuous

In my assignment I have to prove that the following function is discontinuous:

$$f(x)= (2x-1) \space if \space x\notin \Bbb Q $$ $$f(x)=x^2 \space if \space x \in \Bbb Q$$

I have to prove that in terms of $\epsilon$ and $\delta$

I have some sort of beginning, I have to prove that some $\epsilon >0$ exists such that for every $\delta>0$ if $|x-x_0|<\delta \to |f(x)-f(x_0)| >\ \epsilon$

I don't really how to continue from here - I thought about picking $\epsilon=0.01$ and show that $|2x-4|$ is always bigger than that if I pick the right $\delta$ - but how do I show it?

Thanks,

Alan